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Math Help - Points on graphs

  1. #1
    Junior Member Fnus's Avatar
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    Points on graphs

    The two exercises I'd like help with is on the picture, thanks (:
    Attached Thumbnails Attached Thumbnails Points on graphs-1.jpg  
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  2. #2
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    Hello, Fnus!

    Here's the first one . . .


    f(x) \:=\:x^3 + ax + b has a stationary point at (-2,3).

    (a) Find the values of a and b.

    (b) Find the position and nature of all stationary points.
    The point (-2,3) is on the graph.
    . . Hence: . f(\text{-}2) = 3\!:\;\;(\text{-}2)^3 + a(\text{-}2) + b \:=\:3\quad\Rightarrow\quad \text{-}2a + b \:=\:11 .[1]

    f'(x) \:=\:3x^2 + a
    Since (-2,3) is a stationary point, f'(\text{-}2) = 0
    . . and we have: . 3(\text{-}2)^2 + a \:=\:0\quad\Rightarrow\quad\boxed{a \:=\:\text{-}12}

    Substitute into [1]: . \text{-}2(\text{-}12) + b \:=\:11\quad\Rightarrow\quad\boxed{b \:=\:\text{-}13}

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    The function is: . f(x) \:=\:x^3 - 12x - 13

    Then: . f'(x) = 0\quad\Rightarrow\quad 3x^2 - 12 \:=\:0\quad\Rightarrow\quad x^2\:=\:4\quad\Rightarrow\quad x \:=\:\pm2

    When x = 2\!:\;\;f(2) \:=\:2^3 - 12(2) - 13 \:=\:-29 . . . Critical point: (2,\,\text{-}29)

    When x = \text{-}2\!:\;\;f(\text{-}2) \:=\:(\text{-}2)^3 - 12(\text{-}2) - 13 \:=\:3 . . . Critical point: (\text{-}2,\,3)


    Second Derivative Test: . f''(x) \:=\:6x

    f''(2) \:=\:6(2) \:=\:+12 . . . concave up . . . minumum at {\color{blue}(2, \text{-}29)}

    f''\text{(}-2) \:=\:6(\text{-}2) \:=\:-12 . . . concave down . . . maximum at {\color{blue}(\text{-}2, 3)}

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  3. #3
    Junior Member Fnus's Avatar
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    Oh, I see.
    Thanks (:
    That step by step thing makes me actually understand what is going on.
    Thanks a lot!
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Here's 6

    Let P(x) = ax^3 + bx^2 + cx + d

    \Rightarrow P'(x) = 3ax^2 + 2bx + c

    Now, let's piece our clues together.

    Clue 1: P(x) touches the line y = 9x + 2 at the point (0,2)

    we can take two things away from this.

    (1) we interpret the phrase "touching the line" to mean, the line is tangent to the curve. thus it means that P'(0) = 2 \implies \boxed{c = 2}

    (2) the second thing we can take from this, is that P(x) passes through the point (0,2), that is P(0) = 2 which means of course, \boxed{d = 2}

    two unknowns down, two to go.

    what else do we know about P(x)?

    Clue 2: P(x) has a stationary point at (-1,-7)

    again, there are two things we can take from this.

    (1) P(x) passes through (-1,-7), that is, P(-1) = -7

    that means that: a(-1)^3 + b(-1)^2 - 2 + 2 = -7 (i plugged in the values we got for c and d)

    simplifying, we get:

    -a + b = -7 ......................(1)

    (2) the second thing we take from this is that P'(-1) = 0 (that is, the slope of P(x) when x = -1 is zero)

    but that means that: 3a(-1)^2 - 2b - 2 = 0 (i plugged in the value for c)

    simplifying we get:

    3a - 2b = -2 ......................(2)


    thus, to find the remaining unknowns, we must solve the system

    -a + b = -7 ......................(1)
    3a - 2b = -2 ......................(2)

    the solutions are \boxed{a = -16} and \boxed{b = -23}

    thus, P(x) = -16x^3 - 23x^2 + 2x + 2



    EDIT: i made an error. when i concluded that c = 2, i claimed the wrong thing, i should have said that P'(0) = 9 \implies \boxed{c = 9} (since 9 is the slope of the line). make the necessary corrections, i can't bother
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